Theorem undif5 4432

Description: An equality involving class union and class difference. (Contributed by Thierry Arnoux, 26-Jun-2024.)

Assertion

Ref	Expression
undif5	⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∪ 𝐵) ∖ 𝐵) = 𝐴)

Proof of Theorem undif5

Step	Hyp	Ref	Expression
1		difun2 4428	. 2 ⊢ ((𝐴 ∪ 𝐵) ∖ 𝐵) = (𝐴 ∖ 𝐵)
2		disjdif2 4427	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∖ 𝐵) = 𝐴)
3	1, 2	eqtrid 2778	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∪ 𝐵) ∖ 𝐵) = 𝐴)

Syntax hints:   → wi 4   = wceq 1541   ∖ cdif 3894   ∪ cun 3895   ∩ cin 3896  ∅c0 4280

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rab 3396  df-v 3438  df-dif 3900  df-un 3902  df-in 3904  df-nul 4281

This theorem is referenced by:  cyclnumvtx  29778  fressupp  32669  elrgspnlem4  33212  sucdifsn2  38507